# Are the following 2 equations lorentz invariant?

1. Aug 13, 2013

### Dreak

$\partial$μ$\phi$$\partial$μ$\phi$

and

$\partial$μ$\partial$μ$\phi$

with $\phi$(x) a scalar field

2. Aug 13, 2013

### Mentz114

Suppose $\partial_{\mu'}=\Lambda^\mu_{\mu'}\partial_{\mu}$ and $\partial^{\mu'}=\Lambda^{\mu'}_{\mu}\partial^{\mu}$.
Then
$\partial_{\mu'}\partial^{\mu'}\phi=\Lambda^\mu_{\mu'}\partial_{\mu} \Lambda^{\mu'}_{\mu}\partial^{\mu}\phi=\Lambda^\mu_{\mu'} \Lambda^{\mu'}_{\mu}\partial_{\mu}\partial^{\mu}\phi$.

So the scalar is invariant if $\Lambda^\mu_{\mu'}$ is the inverse of $\Lambda^{\mu'}_{\mu}$.

Now, $\Lambda^\mu_{\mu'}$ is not a tensor but a transformation matrix. It so happens that the inverse of the LT is found by reversing the sign of the boost velocity. This operation $\Lambda_b^{b'}=\eta_{ba}\Lambda_{a'}^a\eta^{a'b'}$ also reverses the sign. So for the Lorentz transformation $\Lambda^\mu_{\mu'}$ is the inverse of $\Lambda^{\mu'}_{\mu}$.

3. Aug 13, 2013

### Dreak

I actually thought the second one was lorentzinvariant because:

$\partial$μ$\partial$μ is the d'Alembertian which I thought was Lorentzinvariant, while a scalar field is always Lorentzinvariant?

And what about the other equation? I'm not sure what the influence of the second scalar field in the equation yields.

4. Aug 13, 2013

### Mentz114

The answer is 'yes' for both equations. If you write out the first expression in terms of the transformed coords, there is another cancellation of $\Lambda^\mu_{\mu'}\Lambda_\mu^{\mu'}$.

Rank-0 tensors ( ie no indexes) are always invariant. Differentiating a rank-0 tensor creates a new tensor with rank-1. Scalars formed by contractions of tensors are invariant.

5. Aug 13, 2013

### Dreak

Thank you very much! :)

6. Aug 13, 2013

### vanhees71

One should, however, be a bit more careful with the indices in the Lorentz transformations and write them also with proper order. Let
$${\Lambda^{\mu}}_{\nu}$$
the Lorentz-transformation matrix for contravariant vector/tensor components, i.e.,
$$x'^{\mu}={\Lambda^{\mu}}_{\nu} x^{\nu}.$$
A Lorentz transform obeys
$$\eta_{\mu \nu} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\eta_{\rho \sigma},$$
or
$$\eta_{\mu \nu} \eta^{\rho \alpha} {L^{\mu}}_{\rho} {L^{\nu}}_{\sigma}=\delta_{\sigma}^{\alpha}.$$
This implies
$$\eta^{\rho \alpha} \eta_{\mu \nu} {L^{\mu}}_{\rho}={(L^{-1})^{\alpha}}_{\nu}={L_{\mu}}^{\alpha}.$$

7. Aug 13, 2013

### Mentz114

Thanks for clarifying that.